JEE Advanced 2023 · Paper 1, Mathematics question 8

Rewrite. $\sqrt{1+\cos2x}=\sqrt2\,|\cos x|$, so the equation is
$$|\cos x|=\tan^{-1}(\tan x).$$
On each interval, $\tan^{-1}(\tan x)=x-k\pi$ with $k$ chosen so this lies in $(-\tfrac\pi2,\tfrac\pi2)$.

Put $u=x-k\pi\in(-\tfrac\pi2,\tfrac\pi2)$. Then $|\cos x|=\cos u$, and the equation becomes $\cos u=u$.

Solutions. $\cos u=u$ has exactly one root in $(-\tfrac\pi2,\tfrac\pi2)$, at $u\approx0.739$. It is positive, as it must be since $\cos u>0$.

Each of the three intervals ($k=-1,0,1$) gives exactly one $x$, so there are 3 solutions.